Lecture1 - overview of ODEs#

Overview of ordinary differential equations (ODEs)#

  • linear -> superposition

  • nonlinear -> cannot solve

  • constant coefficients are nice.

First order linear ODEs#

  • general form: $$\frac{dy}{dt} = ay + q(t) \text{ or } \frac{dy}{dt} = a(t)y + q(t)$$

  • rate of change: $$\frac{dy}{dt}$$

Examples of first order ODE.#

solve and assume y(0)=1.

  1. $$\frac{dy}{dt}=y \rightarrow y(t)=e^{t}$$

  2. $$\frac{dy}{dt}=-y \rightarrow y(t)=e^{-t}$$

  3. $$\frac{dy}{dt}=2ty \rightarrow y(t)=e^{t^{2}}$$

  4. $$\frac{dy}{dt}=y^{2} \rightarrow y(t)=\frac{1}{1-t}$$

Second order linear ODEs#

  • second order derivative indicates acceleration or curvature: $$\frac{d^{2}y}{dt^{2}}$$

  • Hooke’s Las: $$\frac{d^{2}y}{dt^{2}} = -ky$$

  • fundamental equation of mechanics $$my’’+by’+ky=f(t),$$ where by’ implies dampling.

System of n equations#

  • first order system, A is a n by n matrix: $$\frac{d\vec{y}}{dt}=A\vec{y}$$

  • second order system, S is a matrix: $$\frac{d^{2}\vec{y}}{dt^{2}}=-S\vec{y}$$

Numerical solutions#

  • MATLAB ode45

  • neural network to solve ODEs

  • ChatGPT to solve ODEs (BE CAREFULL!)

Partial differential euqations (PDEs)#

  • heat equation: $$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}$$

  • wave equation: $$\frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial^{2} u}{\partial x^{2}}$$

  • Laplace equation: $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0$$

The calculus you need#

  • derivitives of functions: $$x^{n}, \sin{x}, \cos{x}, e^{x}, \ln{x}$$

  • rules for derivatives:

    • sum rule

    • product rule

    • quotient rule

    • chain rule: $$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$

  • fundamental theorem of calculus:

$$ \int_{a}^{b} \frac{dy}{dx},dx = y(b) - y(a) $$

$$ \frac{d}{dx}\int_{a}^{x} f(s),ds = f(x) $$

An example, use all rules#

Show that below function y(t) solves $$\frac{dy}{dt}=y+q{t},$$ where q(t) is an external source term.

\[\begin{eqnarray*} y(t) &=& \int_{0}^{t}e^{t-s}q(s)\,ds\\ \Longrightarrow y(t) &=& \int_{0}^{t}e^{t-s}q(s)\,ds = e^{t}\int_{0}^{t}e^{-s}q(s)\,ds\\ \Longrightarrow \frac{dy}{dt} &=& e^{t}\int_{0}^{t}e^{-s}q(s)\,ds + e^{t}(e^{-t}q(t)) = y(t) + q(t) \end{eqnarray*}\]

Tangent line to a graph#

  • Taylor expansion series:

\[\begin{eqnarray*} f(t+\Delta t) \approx f(t)+\Delta t\frac{df}{dt}(t) + \frac{\Delta t^{2}}{2!}\frac{d^{2}f}{dt^{2}}(t)+\cdots + \frac{\Delta t^{n}}{n!}\frac{d^{n}f}{dt^{n}}(t) \end{eqnarray*}\]

Discretized differentiation#

$$\frac{dy}{dx} \approx \frac{\Delta y}{\Delta x}$$

  • approximate y: $$\Delta y \approx \frac{dy}{dx}\Delta x \rightarrow y(x_{o}+\Delta x) \approx y(x_{o})+\Delta x\frac{dy}{dx}(x_{o}), \text{ assuming we know } \frac{dy}{dx}$$

  • Newton’s method (finding roots):

\[\begin{eqnarray*} \Delta x \approx \frac{\Delta y}{\frac{dy}{dx}} \rightarrow x_{1} - x_{o} = \frac{y(x_{1})-y(x_{o})}{\frac{dy}{dx}(x_{o})} \end{eqnarray*}\]

  • central difference

\[\begin{eqnarray*} \frac{dy}{dx}\approx \frac{\Delta y}{\Delta x} = \frac{y(x+\frac{1}{2}\Delta x)-y(x-\frac{1}{2}\Delta x)}{\Delta x} \end{eqnarray*}\]

Solve our first ODE#

\[\begin{eqnarray*} \frac{dy}{dt} = ay, \text{ given } y(0) = y_{o} \end{eqnarray*}\]

We don’t know how to solve, but we can guess the exponential-type solution:

\[\begin{eqnarray*} y(t) = Ae^{\alpha t}, \text{ and so the problem is to find } A \text{ and } \alpha. \end{eqnarray*}\]
\[\begin{eqnarray*} \frac{dy}{dt} = A\alpha e^{\alpha t} \Longrightarrow \alpha = a \Longrightarrow y(t) = Ae^{at} \end{eqnarray*}\]
\[\begin{eqnarray*} y(0) = y_o{} = Ae^{a\times 0} = A \Longrightarrow y(t)=y_{o}e^{at} \end{eqnarray*}\]

  • if a>0, then the solution grows exponentially

  • if a<0, then the solution decays exponentially

Geometry veiw of ODEs#

analytic vs geometry:

\[\begin{eqnarray*} y'(t)=f(x,y) \leftrightarrow \text{ direction field}\\\\ \text{solution } y(x) \leftrightarrow \text{ integral curve} \end{eqnarray*}\]

Superposition#

  • additivity

\[\begin{eqnarray*} F(x_{1}+x_{2}) = F(x_{1}) + F(x_{2}) \end{eqnarray*}\]

  • homogeneity (porportionality a is a scaler)

\[\begin{eqnarray*} F(ax) = aF(x) \end{eqnarray*}\]

  • use additivity and homogeneity, we can derive:

\[\begin{eqnarray*} F(ax_{1}+bx_{2}) = aF(x_{1}) + bF(x_{2}), \end{eqnarray*}\]

which is also called linear combination

Linear ODEs?#

  • are they linear ODEs?

\[\begin{eqnarray*} y'=ky, \\ y'+y^{3}t=y,\\ y'+\cos{x}y=x^{3},\\ \frac{y'}{y}=t^{2},\\ x^{2}yy'+4x=x^{3} \end{eqnarray*}\]

  • show below ODE does not satisify superposition.

\[\begin{eqnarray*} y'+y^{2}=q(t) \end{eqnarray*}\]

Existence and uniqueness#

For the linear first-order ODE, we are not touching the existence, as it is hard to prove consdering the scale of this class. On the other hand, the uniqueness is rather straightforward. Consider

\[\begin{eqnarray*} \frac{dy}{dt} + ay = q(t), y(0)=y_{0} \end{eqnarray*}\]

Assume two solutions are different

\[\begin{eqnarray*} y_{1}(t) \neq y_{2}(t), \forall t \end{eqnarray*}\]
\[\begin{eqnarray*} \frac{y_{1}}{dt}&+&ay_{1}=q(t)\\\\ \frac{y_{2}}{dt}&+&ay_{2}=q(t)\\\\ \end{eqnarray*}\]

Subtract the two equation, we can get

\[\begin{eqnarray*} \frac{d(y_{1}-y_{2})}{dt} + a(y_{1}-y_{2}) = 0 \end{eqnarray*}\]

Let

\[\begin{eqnarray*} Y&=&y_{1}-y_{2}, Y(0)=0\\\\ &\Longrightarrow& \frac{dY}{dt}=-aY\\\\ &\Longrightarrow& Y=e^{-at}, \forall t\\\\ &\Longrightarrow& Y(0) = 0 = A \\\\ &\Longrightarrow& Y(t) = 0, \forall t \\\\ &\Longrightarrow& y_{1}(t) = y_{2}(t), \forall t. \end{eqnarray*}\]

This is a contradiction! So the two solutions have to be the same for all t.

Homogeneous ODE and solution#

Consider

\[\begin{eqnarray*} \frac{dy}{dt} = ay, y(0)=y_{0} \end{eqnarray*}\]

Assume solution is in exponential form

\[\begin{eqnarray*} y(t) = Ae^{\alpha t}. \end{eqnarray*}\]

Plug it into the ODE

\[\begin{eqnarray*} A\alpha e^{\alpha t} = aAe^{\alpha t} \rightarrow \alpha = a \rightarrow y(t)=Ae^{at} \end{eqnarray*}\]

Use initial condition, we can get

\[\begin{eqnarray*} y(0)=y_{0}=Ae^{a\cdot 0}=A \rightarrow y(t) = y_{0}e^{at} \end{eqnarray*}\]

Solution structure#

Consider the ODE with initial condition

\[\begin{eqnarray*} \frac{dy}{dt}=ay+q(t), \text{ given } y(0), \end{eqnarray*}\]

where q(t) is the source term.

We can separate solution into two parts:

  • homogeneous solution or null solution with no source term, which comes from y(0):

\[\begin{eqnarray*} \frac{y_{h}}{dt} = ay_{h}. \end{eqnarray*}\]

  • particular solution, which comes from q(t):

\[\begin{eqnarray*} \frac{y_{p}}{dt} = ay_{p} + q(t) \end{eqnarray*}\]

  • general solution or complete solution

\[\begin{eqnarray*} y(t) = y_{h}(t) + y_{p}(t) \end{eqnarray*}\]